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4y^2-33y+25=0
a = 4; b = -33; c = +25;
Δ = b2-4ac
Δ = -332-4·4·25
Δ = 689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{689}}{2*4}=\frac{33-\sqrt{689}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{689}}{2*4}=\frac{33+\sqrt{689}}{8} $
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